# How to evaluate $\int_{-\infty}^\infty {e^{ax} \over 1 +e^x } \; dx$ [duplicate]

Given that $0 < a < 1$ how to evaluate by the method of residues $$\int_{-\infty}^\infty {e^{ax} \over 1 +e^x } \; dx$$

-
 In particular, see Argon's answer to that question. – Peter Tamaroff Sep 5 '12 at 19:30 @PeterTamaroff how did you change the sum into integral? – hasExams Sep 6 '12 at 4:23 I explain it in detail in the question. – Peter Tamaroff Sep 6 '12 at 13:11

## marked as duplicate by Peter Tamaroff, William, Did, Noah Snyder, J. M.Oct 5 '12 at 12:58

Substituting $u=e^x$, so that $dx = \dfrac{du}{u}$, the integral becomes

$$\int_0^{\infty} \dfrac{u^{a-1}}{1+u}du$$

How might you solve this? You've tagged the question as homework, so I'll leave it here for now, but if you're still stuck, post in the comments.

-
 does this method work when $0 Recalling the$\beta$function, $$\beta( n,m ) = \int_{0}^{1} u^{n-1}\,(1-u)^{m-1}\, du = \frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)} \,.$$ Using the substitution$ 1+u=\frac{1}{t} $casts$ \int_0^{\infty} \dfrac{u^{a-1}}{1+u}du $in terms of the$\beta\$ function,

$$\int_0^{\infty} \frac{u^{a-1}}{1+u}du = \int _{0}^{1}\!{t}^{-a} \left( 1-t \right) ^{a-1}{dt} =\Gamma(a)\Gamma(1-a) = \frac{\pi}{\sin (a \pi )} \,.$$

-
 @Downvoter:I do not know what you are downvoting for. I gave a way to evaluate the integral in a way different from the complex techniques. – Mhenni Benghorbal Sep 5 '12 at 21:04